# Tensor Product with Trivial Vector Space

This question seems obvious and yet I can't seem to find a good answer anywhere.

Let $$V$$ be a finite dimensional vector space, and let $$0$$ denote the trivial vector space. Is $$V \otimes 0 = 0$$ or $$V \otimes 0 = V$$? My gut tells me that it is the second case, but in thinking about dimension, tensor product should multiply dimension in which case I think it is the first case.

• The elements of $V\otimes 0$ are sums of pure tensors of the form ${\bf v}\otimes{\bf 0}$. Then, as we can slide scalars across the $\otimes$ symbol, we have ${\bf v}\otimes{\bf 0}={\bf v}\otimes 0{\bf 0}=0{\bf v}\otimes{\bf 0}={\bf 0}\otimes{\bf 0}$. Thus, there is only one element in $V\otimes 0$, the zero tensor. – runway44 Jul 24 at 22:54
• @runway44 that is IMO a better answer than the two given ones. Why did you post it as a comment? – leftaroundabout Jul 25 at 10:11

Notice that the space of bilinear maps $$f:V\times \{0\}\to k$$ consists of exactly the zero map, therefore the constant map $$w:V\times \{0\}\to\{0\}$$ satisfies the universal property of the tensor product: $$w$$ is bilinear and any bilinear map $$f:V\times\{0\}\to k$$ is the constant zero map, therefore the linear map $$0:\{0\}\to k$$ satisfies $$f=0\circ w$$. $$0$$ is also the only linear map $$\{0\}\to k$$, so it's a fortiori the only linear map $$g$$ such that $$f=g\circ w$$.
Recall that the tensor product $$V\otimes W$$ of two finite-dimensional vector spaces $$V$$ and $$W$$ satisfy the dimension formula $$\dim(V\otimes W)=\dim(V)\cdot\dim(W)$$.
So, tensoring a finite-dimensional vector space $$V$$ with the trivial vector space $$0$$ yields a vector space $$V\otimes 0$$ with dimension $$\dim(V\otimes 0)=\dim(V)\cdot\dim(0)=\dim(V)\cdot 0 = 0$$ This implies that $$V\otimes0$$ is itself the trivial vector space $$V\otimes 0=0$$.
The elements of $$V\otimes 0$$ are sums of pure tensors of the form $${\bf v}⊗{\bf 0}$$. Then, as we can slide scalars across the $$\otimes$$ symbol, we have $${\bf v}\otimes{\bf 0}={\bf v}\otimes (0\cdot{\bf 0})=(0\cdot{\bf v})\otimes{\bf 0}={\bf 0}\otimes{\bf 0}$$.
Thus, there is only one element in $$V\otimes 0$$, the zero tensor.